1990 AHSME Problems/Problem 30

Problem

If $R_n=\tfrac{1}{2}(a^n+b^n)$ where $a=3+2\sqrt{2}$ and $b=3-2\sqrt{2}$ , and $n=0,1,2,\cdots,$ then $R_{12345}$ is an integer. Its units digit is

$\text{(A) } 1\quad \text{(B) } 3\quad \text{(C) } 5\quad \text{(D) } 7\quad \text{(E) } 9$

Solution

$(a+b)R_n=\tfrac12(a^{n+1}+b^{n+1})+ab\cdot\tfrac12(a^{n-1}+b^{n-1})=R_{n+1}+abR_{n-1}$ but $a+b=6$ and $ab=1$ , so this means that $R_{n+1}=6R_n-R_{n-1}$ . Since $R_0=1$ and $R_1=3$ , all terms are integers and we can continue the recurrence $\rm{mod}\ 10$ to get the repeating sequence $1,3,7,9,7,3$ . The number $12345$ is divisible by three but not by six, so the required element in the sequence is $9$ which is answer $\fbox{E}$

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1990 AHSME Problems/Problem 30

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